2011 AMC 12A Problems/Problem 25
Triangle has , , , and . Let , , and be the orthocenter, incenter, and circumcenter of , respectively. Assume that the area of pentagon is the maximum possible. What is ?
Let , , for convenience.
It's well-known that , , and (verifiable by angle chasing). Then, as , it follows that and consequently pentagon is cyclic. Observe that is fixed, hence the circumcircle of cyclic pentagon is also fixed. Similarly, as (both are radii), it follows that and also is fixed. Since is maximal, it suffices to maximize .
Verify that , by angle chasing; it follows that since by Triangle Angle Sum. Similarly, (isosceles base angles are equal), hence Since , by Inscribed Angles.
There are two ways to proceed.
Letting and be the circumcenter and circumradius, respectively, of cyclic pentagon , the most straightforward is to write , whence and, using the fact that is fixed, maximize with Jensen's Inequality.
A more elegant way is shown below.
Lemma: is maximized only if .
Proof by contradiction: Suppose is maximized when . Let be the midpoint of minor arc be and the midpoint of minor arc . Then since the altitude from to is greater than that from to ; similarly . Taking , to be the new orthocenter, incenter, respectively, this contradicts the maximality of , so our claim follows.
With our lemma() and from above, along with the fact that inscribed angles that intersect the same length chords are equal,
Video Solution by Osman Nal
|2011 AMC 12A (Problems • Answer Key • Resources)|
|1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25|
|All AMC 12 Problems and Solutions|